Question

in Fig. 9.25, diagonals AC and BD of quadrilateral
O ABCD intersect at O such that OB = OD.
If AB = CD, then show that:
(i) ar (DOC)=ar (AOB)
(ii) ar (DCB)=ar (ACB)
iii) DA || CB or ABCD is a parallelogram.
(Hint: From D and B, draw perpendiculars to AC.)
this question can be solved without the help of diagram......​

Answer 1

Answer:

Given, ABCD is a quadrilateral. OB = OD and AB = CD .

Const : Draw DM perpendicular AC and BN perpendicular AC .

(i) In triangle DMO and BNO :-

angle 1 = angle 2 ( V.S.O) .............(S)

angle 3 = angle 4 ( each = 90°) .............(A)

OD = OB ( given ) ...............(S)

From SAS cretria :

triangle DMO ≈ triangle BNO

therefore, ar DMO = ar BNO .............. eq(i)

now , in triangle DMC and triangle ABN :-

angle 3 = angle 4 ( each 90°) .........(R)

DM = BN ( by cpct ) .................(H)

AB = CD (given ) ..............(S)

From RHS cretria :

triangle DCM ≈ triangle ABN

Therefore, ar DCM = ar ABN ..............eq(ii)

Adding eq (i) and (ii) :

ar DMO + ar DCM = ar BNO + ar ABN

ar DOC = ar AOB ( H.P) ..................(iii)

(ii) here , triangle BOC is common in both angles .

therefore, adding BOC on both sides of eq (iii) :

ar DOC + ar BOC = ar AOB + ar BOC

ar DCB = ar ABC ( H.P)

(iii) since, DC = AB

And angle DCO = angle CAB ( Proved above )

But these are alternate interior angles

therefore, DC ll AB

And ABCD is a llgm ( H.P )

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